Question

During a car collision, the knee, thighbone, and hip can sustain a force no greater than 4000 N. Forces that exceed this amount could cause dislocations or fractures. Assume that in a collision a knee stops when it hits the car's dashboard. Also assume that the mass of the body parts stopped by the knee is about 20% of the total body mass. The person with a mass of 65 kg is initially traveling at 18 m/s (40 mi/h).

Answer 1

Answer:

a) t = 0.0585 s  

b) s = 0.5265 m

Explanation:

Given:

- maximum possible force F = 4000 N

- mass of person = 65 kg

- mass of knees and other m = 0.2*65 = 13 kg

- initial velocity u = 18 m/s

- Final velocity v = 0

Find:

a) Minimum stopping time

b) Minimum stopping distance.

Solution:

Minimum stopping time can be evaluated by Newton's second law of motion. Where change in momentum per unit time is Average force experienced by the body:

                                   F = m*(v - u) / dt

                                   dt = 13*(0 - (-18)) / 4000

Hence ,                       dt = 0.0585                

       

Minimum stopping distance can be evaluated by using equation of motions and required deceleration is as follows:

                                  a = - 4000  / 13

                                  a = -307.692 m/s^2

And,

                                 s = (v^2 - u^2) / 2*a

                                 s = (0 - 18^2) / 2*(-307.692)

                                s = 0.5265 m  

Answer: The minimum stopping time and distance are t = 0.0585 s and s = 0.5265 m, respectively.